irreducibility of binomials with unity coefficients

If nnn has no oddprime factors, then the binomial xn+1superscriptxn1x^{n}\!+\!1 is irreducible. Thus, x+1x1x\!+\!1, x2+1superscriptx21x^{2}\!+\!1, x4+1superscriptx41x^{4}\!+\!1, x8+1superscriptx81x^{8}\!+\!1 and so on are irreducible polynomials (i.e. irreducible in the fieldℚℚ\mathbb{Q} of their coefficients). N.B., only x+1x1x\!+\!1 and x2+1superscriptx21x^{2}\!+\!1 are irreducible in the field ℝℝ\mathbb{R}; e.g. one has x4+1=(x2-x⁢2+1)⁢(x2+x⁢2+1)superscriptx41superscriptx2x21superscriptx2x21x^{4}\!+\!1=(x^{2}\!-\!x\sqrt{2}\!+\!1)(x^{2}\!+\!x\sqrt{2}\!+\!1).

If nnn is an odd number, then xn+1superscriptxn1x^{n}\!+\!1 is always divisible by x+1x1x\!+\!1:

When nnn is not a prime number but has an odd prime factor ppp, say
n=m⁢pnmpn=mp, then we write xn+1=(xm)p+1superscriptxn1superscriptsuperscriptxmp1x^{n}\!+\!1=(x^{m})^{p}\!+\!1 and apply the idea of (1); for example: